[R] Fitting Pattern-Mixture Models
Dear R-user: Do you know any R package that could be use for fitting a Pattern-Mixture models (PMM) please? I would appreciate if you could suggest me a R package for fitting PMM. Thank you very much, Sincerely, Sattar Got a little couch potato? Check out fun summer activities for kids. idscs=bz [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Embedding Audio Files in Interactive Graphs
Hi Michael Lawrence wrote: On 9/4/07, Sam Ferguson [EMAIL PROTECTED] wrote: Thanks for your reply Bruno. No - as I said, I know how to do that - the movie15 and the multimedia package are basically the same, and it is relatively straightforward to get an audio file into a pdf with them. However, real interactivity is not easily achieved in latex IMO (as it's not its purpose). At least I'm hoping for a bit more flexibility. R seems like a better place to do interactivity, and with the field of information visualisation pointing out that interactivity is a very useful element for investigation of data it seems that clicking around graphical displays may become more and more popular in time. In my field I'm interested in audio data, and so simple interactive visual and auditory displays would be great. A (very useful) start would be 5 separate waveform plots that would play their appropriate sounds when clicked. More complex figures could plot in a 2d space and allow selection of data points or ranges perhaps. I love R for graphics and for Sweave though, and would like to use it if possible - ideally it would be to produce a figure that included the appropriate audiofiles and interactive scripts, which could then be incorporated into a latex document \includegraphics. However, from the deafening silence on this list it seems like I may be attempting to push a square block through a round hole unfortunately. Seems I am back to Matlab and handle graphics - but it won't do this properly either. Lots of things can be embedded into PDF documents, like javascript, flash and svg. Maybe it would be feasible to use the gridSVG package to output some graphics as svg with javascript to play the sounds and embed that into a pdf? The short answer is that R cannot do this sort of thing and is unlikely to be able to do it anytime soon. The basic problem is that core R graphics has no concept of animation, audio, hyperlinks, etc AND there is no way to access these features on devices that do have these concepts (e.g., PDF). It would be nice to change that, but it is a large redesign problem that is not anywhere near the top of anyone's todo list (to my knowledge). As Michael mentioned, the gridSVG package allows you to draw using the grid package and access some of the fancier SVG features (including embedding scripts). It has its own problems of course, but may be worth trying. Paul Cheers Sam On 03/09/2007, at 5:39 PM, Bruno C.. wrote: Are you asking on how to include an audio file into a pdf? This is already feasible via latex and the movie 15 package ;) Ciao Hi R-ers, I'm wondering if anyone has investigated a method for embedding audio files in R graphs (pdf format), and allowing their playback to be triggered interactively (by clicking on a graph element for instance). I know how to do this in latex pdfs with the multimedia package, but it seems that R would provide a more appropriate platform for many reasons. Thanks for any help you can provide. Sam Ferguson Faculty of Architecture, Design and Planning The University of Sydney __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Leggi GRATIS le tue mail con il telefonino i-mode™ di Wind http://i-mode.wind.it/ -- Sam Ferguson Faculty of Architecture The University of Sydney [EMAIL PROTECTED] +61 2 93515910 0410 719535 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [Lattice] Incrase the height of strips in Trellis plots
This should give you something close to what you want: xyplot(Petal.Length ~ Petal.Width | Species, iris, strip = strip.custom(par.strip.text = list(cex = 2)), par.settings = list(layout.heights=list(strip=1.45))) The par.settings argument alters locally the default par settings of lattice plots, see e.g. ?trellis.par.get and the ?xyplot about the par.settings argument. A closer inspection (try using strip = 2 or numbers less than 1.45 in stead of strip = 1.45) of the figure reveals that there are some problems with vertical alignment of the strip text, i.e. not centered. To remedy this I think that you have to use you own strip function. You may be able to build you own function by altering the default strip function, see ?lattice.options and try lattice.options() at the command prompt. However this may be more difficult than it seems at first. Perhaps some one more familiar with the lattice package can solve this. Best regards Frede Aakmann Tøgersen Scientist UNIVERSITY OF AARHUS Faculty of Agricultural Sciences Dept. of Genetics and Biotechnology Blichers Allé 20, P.O. BOX 50 DK-8830 Tjele Phone: +45 8999 1900 Direct: +45 8999 1878 E-mail: [EMAIL PROTECTED] Web: http://www.agrsci.org This email may contain information that is confidential. Any use or publication of this email without written permission from Faculty of Agricultural Sciences is not allowed. If you are not the intended recipient, please notify Faculty of Agricultural Sciences immediately and delete this email. -Oprindelig meddelelse- Fra: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] På vegne af Bernd Weiss Sendt: 5. september 2007 06:48 Til: r-help@stat.math.ethz.ch Emne: [R] [Lattice] Incrase the height of strips in Trellis plots Dear all, I wonder how to increase the height of strips via strip.default or strip.custom. The following example hopefully illustrates the difficulty I am facing: library(lattice) xyplot(Petal.Length ~ Petal.Width | Species, iris, strip = strip.custom(par.strip.text = list(cex = 2))) Thanks for any advice, Bernd version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 5.1 year 2007 month 06 day27 svn rev42083 language R version.string R version 2.5.1 (2007-06-27) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to extract part of an array?
Hi, How can I extract part of an array? I would like to extract table Supported from this array. If this is not possible, how do I convert array to list? I'm sorry this is not an reproducible example. spl - tapply(temp$var1, list(temp$var2, temp$var3, temp$var3), mean) spl , , Supported 07 08 A68.38710 71.48387 B21.67742 20.83871 C55.74194 61.12903 AL L 26.19816 27.39631 , , Not_supported 07 08 ANA 82.38710 BNA 24.0 CNA 68.77419 ALL NA 29.97984 -Lauri __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] capture.out(system())?
On 9/4/07, Werner Wernersen [EMAIL PROTECTED] wrote: Hi, I am trying to capture the console output of program I call via system() but that always returns only character(0). For example: capture.output(system(pdflatex out.tex) ) will yield: character(0) and the output still written to the R console. Is there a command for intercepting this output? Thank you! Werner ?sink() -- Gustaf Rydevik, M.Sci. tel: +46(0)703 051 451 address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [Lattice] Incrase the height of strips in Trellis plots
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Frede Aakmann Tøgersen schrieb: This should give you something close to what you want: xyplot(Petal.Length ~ Petal.Width | Species, iris, strip = strip.custom(par.strip.text = list(cex = 2)), par.settings = list(layout.heights=list(strip=1.45))) The par.settings argument alters locally the default par settings of lattice plots, see e.g. ?trellis.par.get and the ?xyplot about the par.settings argument. A closer inspection (try using strip = 2 or numbers less than 1.45 in stead of strip = 1.45) of the figure reveals that there are some problems with vertical alignment of the strip text, i.e. not centered. To remedy this I think that you have to use you own strip function. You may be able to build you own function by altering the default strip function, see ?lattice.options and try lattice.options() at the command prompt. However this may be more difficult than it seems at first. Perhaps some one more familiar with the lattice package can solve this. Perfect! Thank you very much, Bernd -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.7 (MingW32) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org iD8DBQFG3l5cUsbvfbd00+ERAh1qAJ4tw3ZiIYnI+UF6FJeLT1xRMep/VACfR+I1 R14RJkdaFBNFqVc6kibyaRk= =W41I -END PGP SIGNATURE- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] question about non-linear least squares in R
Hi, everyone, My question is: It's not every time that you can get a converged result from the nls function. Is there any solution for me to get a reasonable result? For example: x - c(-0.06,-0.04,-0.025,-0.015,-0.005,0.005,0.015,0.025,0.04,0.06) y - c(1866760,1457870,1314960,1250560,1184850,1144920,1158850,1199910,1263850,1452520) fitOup- nls(y ~ constant + A*(x-MA)^4 + B*(x-MA)^2, start=list(constant=1000, A=1, B=-100, MA=0), control=nls.control(maxiter=100, minFactor=1/4096), trace=TRUE) For this one, I cannot get the converged result, how can I reach it? To use another funtion or to modify some settings for nls? Thank you very much! Yours, Warren __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [Lattice] Incrase the height of strips in Trellis plots
On 9/5/07, Frede Aakmann Tøgersen [EMAIL PROTECTED] wrote: This should give you something close to what you want: xyplot(Petal.Length ~ Petal.Width | Species, iris, strip = strip.custom(par.strip.text = list(cex = 2)), par.settings = list(layout.heights=list(strip=1.45))) The par.settings argument alters locally the default par settings of lattice plots, see e.g. ?trellis.par.get and the ?xyplot about the par.settings argument. Another possibility that predates par.settings (but is basically equivalent) is xyplot(Petal.Length ~ Petal.Width | Species, iris, par.strip.text = list(lines = 2, cex = 2)) -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Robust linear models and unequal variance
Thanks for your reply, Bert - it was really helpful! 1. I have used a non-parametric test of variance (fligner.test) to test whether the variances of my groups were equal or not, and they weren't. 2. Just quickly to see if I understand the weighting part - I calculate the variance of my different group (I have two factors, so I presume calculating it for all different combinations of these two factors separately) and use a weight 1/variance for each of these groups in the GLM. Right? 3. OK 4. Unfortunately my group sizes are dissimilar, with most groups being approximately equal sample size and one group having a much higher sample size - so best thing to do is use weights than? Thanks again, Geertje Geertje van der Heijden PhD student Tropical Ecology School of Geography University of Leeds Leeds LS2 9JT Tel: (+44)(0)113 3433345 Email: [EMAIL PROTECTED] -Original Message- From: Bert Gunter [mailto:[EMAIL PROTECTED] Sent: 04 September 2007 23:24 To: Geertje Van der Heijden; r-help@stat.math.ethz.ch Subject: RE: [R] Robust linear models and unequal variance Let me try a reply, although I wish others wiser than I had responded. 1. How do you know the variances are unequal? 2. If you somehow know what the variances are (or at least their relative sizes), you can use the weights arguments of the functions you mentions to weight inversely proportional to variance (except not for the MM method in rlm() according to the docs.) 3. That ranked regression is robust is a myth. It also does not deal with the unequal variance situation. It is not a panacea for anything. If you need robust regression use robust regression. 4. If group sizes are not too dissimilar, than whether you case weight or not may not make much difference (alas, hard to tell a priori). Especially to estimation. The fundamental issue is that outliers and unequal variances must be operationalized, otherwise they are confounded: outlier only has meaning compared to what is expected from a specified distribution. Outliers are no longer out when the variance is large. Also look at glm() with the quasi option if you wish to consider fitting a heterogeneous variance structure to initialize a robust method (which could, of course, be distorted by your outliers). Bert Gunter Genentech Nonclinical Statistics -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Geertje Van der Heijden Sent: Tuesday, September 04, 2007 10:55 AM To: r-help@stat.math.ethz.ch Subject: [R] Robust linear models and unequal variance Hi all, I have probably a basic question, but I can't seem to find the answer in the literature or in the R-archives. I would like to do a robust ANCOVA (using either rlm or lmRob of the MASS and robust packages) - my response variable deviates slightly from normal and I have some outliers. The data consist of 2 factor variables and 3-5 covariates (fdepending on the model). However, the variance between my groups is not equal and I am not sure if it is therefore appropriate to use a robust statistical method or if a non-parametric analysis (i.e. ranked regression) might be better. If I can still use a robust statistical method, which estimator is best to use to deal with unequal variance? And if it is better to use a non-parametric analysis, could anyone put me in the direction of the right non-parametric method to use (the relationship between my response variable and the covariates is linear)? Any help on this would be greatly appreciated! Many thanks, Geertje Geertje van der Heijden PhD student Tropical Ecology School of Geography University of Leeds Leeds LS2 9JT Tel: (+44)(0)113 3433345 Email: [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] capture.out(system())?
On Wed, 5 Sep 2007, Gustaf Rydevik wrote: On 9/4/07, Werner Wernersen [EMAIL PROTECTED] wrote: Hi, I am trying to capture the console output of program I call via system() but that always returns only character(0). For example: capture.output(system(pdflatex out.tex) ) will yield: character(0) and the output still written to the R console. Is there a command for intercepting this output? Thank you! Werner ?sink() That is used by capture.output() to capture R output, but this question is about output that never goes near R. The answer is in ?system, but might depend on the unstated OS. Arguments 'intern' and 'show.output.on.console' are relevant. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] question about non-linear least squares in R
Below is one possibility: If you knew MA you would get a regular linear least-squares for parameters A,B and constant which can be easily solved. So now you can define a function f(MA) which returns that value. Now you must minimize that f - a function of one argument. It can have several local minima and so you must be careful but I believe that minimizing (even bad) function of one argument should be easier than your original problem. Regards, Moshe. P.S. if you do this I would be interested to know whether this works. --- Yu (Warren) Wang [EMAIL PROTECTED] wrote: Hi, everyone, My question is: It's not every time that you can get a converged result from the nls function. Is there any solution for me to get a reasonable result? For example: x - c(-0.06,-0.04,-0.025,-0.015,-0.005,0.005,0.015,0.025,0.04,0.06) y - c(1866760,1457870,1314960,1250560,1184850,1144920,1158850,1199910,1263850,1452520) fitOup- nls(y ~ constant + A*(x-MA)^4 + B*(x-MA)^2, start=list(constant=1000, A=1, B=-100, MA=0), control=nls.control(maxiter=100, minFactor=1/4096), trace=TRUE) For this one, I cannot get the converged result, how can I reach it? To use another funtion or to modify some settings for nls? Thank you very much! Yours, Warren __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Choosing the optimum lag order of ARIMA model
Hi Leeds, Thanx for this reply. Actually I did not want to know whether any differentiation is needed or not. My question was that : what is the difference between two models : arima(data, c(2,1,2)) and arima(diff(data), c(2,0,2)) If I am correct then those two models are same. Therefore I should get same results for both of the cases. Am I doing something wrong? Leeds, Mark (IED) [EMAIL PROTECTED] wrote: you shouldn't just do a diff because the non diffed version gives you an error. I don't know what that error means but you definitely can't just ignore it and go to taking a difference. Why don't you do an acf plot of the non diffed series and see if the acf doesn't die out quickly. If it doesn't, then it's probably okay to assume you need to difference it. if you check out the source of the function, that might gives hintsa about what the error means. Whayt you say below about looking at combinations is okay but remember that picking a model is an art rather than S science. Maybe an arima(2,1,2) is the best model based on model selection and aic but it gives forecasts that Are very poor. Parsimony ( fewer parameters ) is stressed by boix and jenkins so, when in doubt,, choose a lower order model when all else fails. The series may not have an perfect arima represenation so nothing is going to be perfect. -Original Message- From: Megh Dal [mailto:[EMAIL PROTECTED] Sent: Saturday, September 01, 2007 1:20 AM To: Leeds, Mark (IED) Cc: r-help@stat.math.ethz.ch Subject: RE: [R] Choosing the optimum lag order of ARIMA model Hi Leed, I got your point. Hence if I see both acf and pacf vanish after 3 then I should try for all possible models and then choose that model giving min aic? i.e. (1,3), (3,1), (3,3), (2,3), (3,2), (1,2), (2,1), (1,1), and (2,2)? And my second doubt is : for the particular dataset that I provided, I got nothing when I run arima(data, order=c(2,1,2)) however arima(diff(data), order=c(2,0,2)) gives no problem : arima(data, order=c(2,1,2)) Error in arima(data, order = c(2, 1, 2)) : non-stationary AR part from CSS arima(diff(data), order=c(2,0,2)) Call: arima(x = diff(data), order = c(2, 0, 2)) Coefficients: ar1 ar2 ma1 ma2 intercept 0.1093 -0.3111 -0.1438 0.0632 0.0157 s.e. 0.5378 0.4464 0.5661 0.4796 0.0111 sigma^2 estimated as 0.01329: log likelihood = 47.38, aic = -82.76 Can anyone tell me what is the wrong there? Regars, Leeds, Mark (IED) wrote: what ripley says below is kind of related to what I said about p and q both being greater than 1 being very unlikely. He's also right in that those rules only work in the sense that, if the acf drops off after q lags, then the Implication is that p = 0 And if they pacf drops off after p lags, then it's implied that q = 0. when the model is mixed, it's more complicated and Mixed models are more rare than common but they could end up being the best model. That's another place where The aic can be used. In other words, if it looks like your acf drops off after 1 and your pacf drops off after 1, then it could be a p = 1 and q =1 model but then the aic should be checked against ( p =1 and q = 0 ) And p = 0 and q = 1 ) because the selection of p = 1 and q = 1 is really flawed because the rules don't really Hold when BOTH p and q are non zero. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Prof Brian Ripley Sent: Friday, August 31, 2007 4:38 AM To: Megh Dal Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Choosing the optimum lag order of ARIMA model On Fri, 31 Aug 2007, Megh Dal wrote: Dear all R users, I am really struggling to determine the most appropriate lag order of ARIMA model. My understanding is that, as for MA [q] model the auto correlation coeff vanishes after q lag, it says the MA order of a ARIMA model, and for a AR[p] model partial autocorrelation vanishes after p lags it helps to determine the AR lag. And most appropriate model choosed by this argument gives min AIC. The last part is fallacious. Also, you are applying your rules to selecting the orders in ARMA models, and they apply only to pure MA or AR models. The R test file src/library/stats/tests/ts-tests.R has an example of model selection by AIC. Now I considered following data : 2.1948 2.2275 2.2669 2.2839 1.9481 2.1319 2.0238 2.3109 2.5727 2.5176 2.5728 2.6828 2.8221 2.879 2.8828 2.9955 2.9906 2.9861 3.0452 3.068 2.9569 3.0256 3.0977 2.985 2.9572 3.0877 3.1009 3.1149 2.8886 2.9631 3.0325 2.9175 2.7231 2.7905 2.8493 2.8208 2.8156 2.9115 2.701 2.6928 2.7881 2.723 2.7266 2.9494 3.113 3.0566 3.0358 3.05 3.0724 3.1365 3.1083 3.0257 3.2211 3.4269 3.327 3.1205 2.9997 3.0201 3.0803 3.2059 3.1997 3.038 3.1613 3.2802 3.2194 ACF for 1st diff series: Autocorrelations of series 'diff(data1)', by lag 0 1 2 3 4 5 6 7 8 9 10 1.000 -0.022 -0.258 -0.016 0.066 0.034 0.035 -0.001 -0.089 0.028 0.222 11 12 13 14 15 16 17 18 -0.132 -0.184 -0.038 0.048 -0.026 -0.041
Re: [R] Pie Chart Labels
Adam Green wrote: I am having trouble finding out how to adjust the position of labels on pie charts. For the small wedges, many of the labels overlap making it impossible to read. Is there any way to offset the labels so that they don't overlap? Hi Adam, There are three ways to adjust the positions of labels on a 3D pie chart (I'm assuming it's 3D from the word wedges. If not, the following should still be useful). The first is to change the position of the sectors by the start argument. Apart from being useful when very large sectors cause problems, changing the sector positions can sometimes alleviate minor crowding of the labels. The second is to catch the return value of the pie3D function. This is the radial positions at which the labels have been placed. You can then alter this vector and pass it as the argument labelpos. In this way you can spread out the labels for small sectors. Finally, you can leave out the labels when you call pie3D and manually place labels using the text function. The chart is drawn on a plot that spans -1 to 1 in both directions, so it is relatively easy to work out where the labels should be. A word of warning - when I read, For the small wedges, many of the labels..., I felt that I should say that pie charts in general, and 3D ones in particular, are rarely successful with more than 4 or 5 sectors. In the R-news article that included pie3D, I included a pie chart that was very difficult for the viewer to extract the information that the creator intended. Depending upon what is to be represented, you might consider whether another method might be more successful. Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to extract part of an array?
Lauri Nikkinen wrote: Hi, How can I extract part of an array? I would like to extract table Supported from this array. If this is not possible, how do I convert array to list? I'm sorry this is not an reproducible example. spl - tapply(temp$var1, list(temp$var2, temp$var3, temp$var3), mean) spl , , Supported 07 08 A68.38710 71.48387 B21.67742 20.83871 C55.74194 61.12903 AL L 26.19816 27.39631 , , Not_supported 07 08 ANA 82.38710 BNA 24.0 CNA 68.77419 ALL NA 29.97984 How about spl[,,Supported]? -p -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice xyplot with bty=l
Deepayan Sarkar [EMAIL PROTECTED] writes:
On 9/4/07, Patrick Drechsler [EMAIL PROTECTED] wrote:
what is the correct way of removing the top and right axes
completely from a lattice xyplot? I would like to have a plot similar
to using the bty=l option for traditional plots.
There is no direct analog (and I think it would be weird in a
multipanel plot).
I agree that this is not very useful for multipanel plots.
Combining a few different features, you can do:
library(grid)
xyplot(1:10 ~ 1:10, scales = list(col = black, tck = c(1, 0)),
par.settings = list(axis.line = list(col = transparent)),
axis = function(side, ...) {
if (side == left)
grid.lines(x = c(0, 0), y = c(0, 1),
default.units = npc)
else if (side == bottom)
grid.lines(x = c(0, 1), y = c(0, 0),
default.units = npc)
axis.default(side = side, ...)
})
-Deepayan
Thank you very much Deepayan, this is exactly what I was looking for!
Cheers,
Patrick
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Re: [R] integers
On 9/4/07, Christoph Heibl [EMAIL PROTECTED] wrote: Hi list, The function is.integer tests if an object is of type integer: see e.g.: is.integer(12) # FALSE is.real(12) # TRUE mode(12)# numeric But how can I test if a number is actually an integer? R seek is difficult to search in this case because it mainly yields entries about the integer()-function family. Thanks for any hint! Christoph Heibl ?mode r = 12 is.integer(r) [1] FALSE is.double(r) [1] TRUE i = as.integer(r) storage.mode(r) [1] double storage.mode(i) [1] integer Paolo Sonego __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Running geeglm unstructured corstr
Dear R users, I've got a serious problem running some gee functions, and I really can't fix it. My dataframe is made of several rows and columns (say 7600 x 15), like the one below: header inr . inside group_cod 1 2.25 0 1 1 3.46 0 0 1 .. 1 0 1 .. 1... 1 .. . ... .. . ... .. . ... .. . ... .. . ... .. . ... 615 .. . ... 615 .. . ... 615 .. . ... As you can see I've got several repeated measures, resulting in clusters ( i.e. ID) of different sizes. You can get a frame like this by running this code: header - rep.int(seq(1:615),sample(seq(1:19),size=615,replace=T)) inr - rlnorm(length(header),0.8434359,0.3268392) group_cod - sample(c(0,1),size=length(header),replace=T) inside - sample(c(0,1),size=length(header),replace=T) gee.frame - data.frame(header,inr,group_cod,inside) When I try running a longitudinal model with geeglm, the corstr=unstructured option returns me an assertion failure of R itself (a sort of error window), while the ar1 option for corstr returns a model only with estimates, and no s.e. or wald test. Same is for the anova method. I removed subjects with only one observation, but the problem's still the same. This is the code used: - for corstr= unstructured geeglm.model- geeglm(inside~group_cod,family=binomial, data=gee.frame ,id=header,corstr=unstructured) This model gives me an error (not the kind written in R-gui: simply the program seems to stop not producing anything), and doesn't produce any result, leading me to quit R. The same happens when I use a userdefined matrix built with genZcor function using crostrv=4 (i.e. unstructured). My initial thought was of problems because of highly unbalanced observations and because of exceeding of correlation parameters to estimate, but restricting cluster size did not result in any improvement. Is there a problem with the code, or it could be due to the data? - for costr= ar1 In my dataset I also have a variable weeks, that I use to specify waves in the geeglm. When I use this, the output for an autoregressive gee model gives me only the estimates, when I don't use it everything's alright. The problem's that I have unbalanced observations, so the use of waves could take this into account while estimating parameters. Is this a conflict between these options, or what? In addition, I've built the empirical correlation structure. A 19x19 structure, thus with nrow=maximum cluster size. When I use this as zcor the program gives me geeglm.fixed - geeglm(inside~weeks+group_cod+età,family=binomial, data= frame.model,id=header,waves=weeks,zcor=corr.gee,corstr=userdefined) *Errore in geese.fit(xx, yy, id, offset, soffset, w, waves = waves, zsca, : nrow(zcor) need to be equal sum(clusz * (clusz - 1) / 2) for unstructured or userdefined corstr.* I really don't understand the meaning of this: the correlation should have number of rows equal to maximum cluster size. Also in the online help the details say something about the dimension of Zcor, but I can't understand the same... I hope I've been clear. Thanks in advance niccolò [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Choosing the optimum lag order of ARIMA model
On Wed, 5 Sep 2007, Megh Dal wrote: Hi Leeds, Thanx for this reply. Actually I did not want to know whether any differentiation is needed or not. My question was that : what is the difference between two models : arima(data, c(2,1,2)) and arima(diff(data), c(2,0,2)) If I am correct then those two models are same. Therefore I should get same results for both of the cases. Am I doing something wrong? They are not the same. Please do study the help page, and in particular the 'include.mean' argument. One is a model for n observations and the other for n-1 observations, and how that affects the issue is discussed on the help page. With the right options you will get similar but not identical results. arima(x, c(2,1,2), method=ML) Coefficients: ar1 ar2 ma1 ma2 0.0786 -0.3561 -0.0869 0.1272 s.e. 0.6135 0.4296 0.6564 0.4549 sigma^2 estimated as 0.01368: log likelihood = 46.46, aic = -82.92 arima(diff(x), c(2,0,2), method=ML, include.mean=FALSE) Coefficients: ar1 ar2 ma1 ma2 0.0786 -0.3561 -0.0869 0.1272 s.e. 0.6135 0.4296 0.6564 0.4549 sigma^2 estimated as 0.01329: log likelihood = 47.38, aic = -82.76 And did you have permission to copy private (and impolite) messages from Mr Leeds to this list? If you did, please say so in your own posting for the record. Since I don't have such permission I have deleted them from this reply. Professor Ripley -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plotting
This looks like exactly what i want!! Thanks! Jim Lemon-2 wrote: yoo wrote: Hi, let's say I have data x = c(1, 2, 10, 12) y = c(100, -20, 50, 25) if I go plot(x, y), then the default x-axis range goes from 1 to 12. Is there a way to change it so that the axis looks like: |-|-|-| 1 2 10 12 This doesn't seem reasonable but let's say I want to plot intraday graph with axis.POSIXct, my data is only from 8:30 to 4 every day and I have these data for days.. i don't want to see a straight line every night.. Is there a way I can do this? Hi yoo, Have you looked at gap.plot (plotrix)? Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/plotting-tf4361709.html#a12497565 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Different behavior of mtext
You may also want to look at the cnvrt.coords function in the TeachingDemos
package. It may be a bit simpler than mixing grid and base.
-Original Message-
From: Sébastien [EMAIL PROTECTED]
To: Prof Brian Ripley [EMAIL PROTECTED]
Cc: R-help r-help@stat.math.ethz.ch
Sent: 9/3/07 7:46 PM
Subject: Re: [R] Different behavior of mtext
Thanks for the information on gridBase, I could solve my problem using
the 'baseViewports' function and by replacing mtext by grid.text (with
coordinates adjustments).
Sebastien
Prof Brian Ripley a écrit :
On Mon, 3 Sep 2007, Sébastien wrote:
Ok, the problem is clear now. I did not get that 'user-coordinates'
was refering to par(usr), when I read the help of mtext. If I may
ask you some additional questions:
- you mentioned a missing unit() call ; at which point should it be
done in my code examples ?
Before it is used. The problem is that I believe more than one
package has a unit() function.
- could you give me some advices or helpful links about how to set up
a grid viewport ? - and finally, probably a stupid question: is a
gridview automatically set up when a plotting function is called ?
If you want to mix grid and base graphics, you need package gridBase,
but really I would not advise a beginner to be using grid directly
(that is, not via lattice to ggplot*).
Sebastien
PS: To answer to your final question, my goal is to center a block of
legend text on the device but to align the text to the left of this
block.
Prof Brian Ripley a écrit :
On Sun, 2 Sep 2007, Sébastien wrote:
Dear R Users,
I am quite surprised to see that mtext gives different results when it
is used with 'pairs' and with plot'. In the two following codes, it
seems that the 'at' argument in mtext doesn't consider the same
unit system.
It is stated to be in 'user coordinates'. Your code does not work
because unit() is missing. If you mean the one from package grid,
npc is not user coordinates (and refers to a grid viewport which
you have not set up and coincidentally is the same as the initial
user coordinate system to which pairs() has reverted).
Try par(usr) after your pairs() and plot() calls to see the
difference.
Plotting a 2x2 array of plots _is_ different from plotting one, so
this should be as expected.
Since centring is the default for 'adj', it is unclear what you are
trying to achieve here.
I would appreciate your comments on this issue.
Sebastien
# Pairs
mydata-data.frame(x=1:10,y=1:10)
par(cex.main=1, cex.axis=1, cex.lab=1, lwd=1,
mar=c(5 + 5,4,4,2)+0.1)
pairs(mydata,oma=c(5 + 5,4,4,2))
mylegend-c(mylegend A,mylegend B,mylegend C,mylegend test)
mylegend.width = strwidth(mylegend[which.max(nchar(mylegend))],
figure)
for (i in 1:4) {
mtext(text=mylegend[i],
side = 1,
line = 3+i,
at = unit((1-mylegend.width)/2,npc),# centers the
legend at the bottom
adj=0,
padj=0)}
# plot
mydata-data.frame(x=1:10,y=1:10)
par(cex.main=1, cex.axis=1, cex.lab=1, lwd=1,
mar=c(5 + 5,4,4,2)+0.1)
plot(mydata,oma=c(5 + 5,4,4,2))
mylegend-c(mylegend A,mylegend B,mylegend C,mylegend test)
mylegend.width = strwidth(mylegend[which.max(nchar(mylegend))],
figure)
for (i in 1:4) {
mtext(text=mylegend[i],
side = 1,
line = 3+i,
at = unit((1-mylegend.width)/2,npc),# should
center the legend at the bottom but doesn't do it !
adj=0,
padj=0)}
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Re: [R] Help: how can i build a constrained non-linear model?
Dear all I try to run the code as follows, test.model-nls(y~exp(A)*(x-PMA)^4+exp(B)*(x-PMA)^2+Const, data=test, start=list(A=8,B=5,Const=10,PMA=0), control=nls.control(maxiter = 50,minFactor=1/1048), trace=TRUE) But how can i build a selfSart, since i have much data ? Thanks for your help first! Vina From: [EMAIL PROTECTED]: [EMAIL PROTECTED]: Help: how can i build a constrained non-linear model?Date: Tue, 4 Sep 2007 07:53:41 + DearI have a data.frame, and want to fit a constrained non-linear model:data: x y -0.08 20.815 -0.065 19.8128 -0.05 19.1824 -0.03 18.7346 -0.015 18.3129 0.015 18.0269 0.03 18.4715 0.05 18.9517 0.065 19.4184 0.08 20.146 0 18.2947model:y~exp(a)*(x-m)^4+exp(b)*(x-m)^2+const I try to use nls() and set start=list(a=1,b=1,c=1,m=1), but which always give me a error message that Error in qr.solv(QR.B,cc): singular matrix 'a' in solve. How can i build a selfStart? or any suggestion! ThanksRegards,Vina Invite your mail contacts to join your friends list with Windows Live Spaces. It's easy! Try it! _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Spline
I want to interpole a functione by a spline interpolation and i want to know if i can force, with any function or parameter, the interpolation to be monotonic. Thanks all -- View this message in context: http://www.nabble.com/Spline-tf4384168.html#a12498336 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] length of a string
Dear all, I would like to know how can I compute the length of a string in a dataframe. Example: SEQUENCE ID TGCTCCCATCTCCACGGHR04FS00645 ACTGAACTCCCATCTCCAAT HR0595847847 I would like to know how to compute the length of each SEQUENCE. Best regards, João Fadista [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] length of a string
sLengths - with(dataFrame, nchar(as.character(SEQUENCE))) Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of João Fadista Sent: Wednesday, 5 September 2007 11:51 PM To: r-help@stat.math.ethz.ch Subject: [R] length of a string Dear all, I would like to know how can I compute the length of a string in a dataframe. Example: SEQUENCE ID TGCTCCCATCTCCACGGHR04FS00645 ACTGAACTCCCATCTCCAAT HR0595847847 I would like to know how to compute the length of each SEQUENCE. Best regards, João Fadista [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] For loop with if else statement
Hans,
I think your problem is that you don't use the variable which takes different
values in your if statement your i changes values and has really nothing
to do with your x variable (except the length part ). Also all the other
variables need to be declared somehow - otherwise how can you store values in
them???
So your first code may be something like that:
t=c(1,2)
for(i in 1:length(t)){
if (i==1) t[i]=i+1 else t[i]=i
}
t[1] 2 2
and your second code:
a = c(1,2)
b=c(1,2)for(i in 1:2){if (i==1){a[i]=ib[i]=i-1} else{
a[i]=i+1b[i]=i}c-list(a=a,b=b)}
c$a[1] 1 3 c$b[1] 0 2
Also when you have only 2 possible values for i i don't think the second if
is necessary. I hope this helps a little, although my explanation is not
necessarily the best.
Monica
-- Message: 16Date: Tue, 4 Sep 2007 15:59:54
+0200From: Hans Ole ?rka [EMAIL PROTECTED]Subject: [R] For loop with if else
statementTo: 'r-help@stat.math.ethz.ch'
r-help@stat.math.ethz.chMessage-ID:[EMAIL PROTECTED]Content-Type:
text/plain; charset=iso-8859-1 Hi,I try to make a simple for loop with a if
else statement (First example - Below) and extend it to a more complex loop
(Second example). However, my results #First example:x=c(1,2)t=for(i in
1:length(x)){if (x==1){a=x+1}elseif (x==2){a=x}} Returned from R:Warning
messages:1: the condition has length 1 and only the first element will be
used in: if (x == 1) {2: the condition has length 1 and only the first
element will be used in: if (x == 1) { t[1] 2 3 However, the result i had
liked to get was t=c(2,2) i.e. using the first function (a=x+1) for x[1] and
(a=x) for x[2]. I can remove the Warnings by making: if (x[i]==1) etc. but this!
do not make the results any better. #Second example:x=c(1,2)t-for(i in
1:length(x)){if (x==1){a=xb=x-1}elseif (x==2){a=x+1b=x}b-list(a=a,b=b)}
Returned from R:Warning messages:1: the condition has length 1 and only the
first element will be used in: if (x == 1) {2: the condition has length 1 and
only the first element will be used in: if (x == 1) { t$a[1] 1 2 $b[1] 0 1 The
result i like to get are $a =c(1,3) and $b=c(0,2) Probably there are couple of
things that I do wrong and I appreciate all help!
_
Discover the new Windows Vista
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] length of a string
How's this? x = data.frame(ID=c(asdf,asdfasdf),1:2) x ID X1.2 1 asdf1 2 asdfasdf2 nchar(as.character(x$ID)) [1] 4 8 Assuming ID is a factor, if not, you can remove the as.character(). On 9/5/07, João Fadista [EMAIL PROTECTED] wrote: Dear all, I would like to know how can I compute the length of a string in a dataframe. Example: SEQUENCE ID TGCTCCCATCTCCACGGHR04FS00645 ACTGAACTCCCATCTCCAAT HR0595847847 I would like to know how to compute the length of each SEQUENCE. Best regards, João Fadista [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] length of a string
João Fadista wrote: Dear all, I would like to know how can I compute the length of a string in a dataframe. Example: SEQUENCE ID TGCTCCCATCTCCACGGHR04FS00645 ACTGAACTCCCATCTCCAAT HR0595847847 I would like to know how to compute the length of each SEQUENCE. ?nchar Best regards, João Fadista [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] length of a string
On 9/5/07, João Fadista [EMAIL PROTECTED] wrote: I would like to know how can I compute the length of a string in a dataframe. Example: SEQUENCE ID TGCTCCCATCTCCACGGHR04FS00645 ACTGAACTCCCATCTCCAAT HR0595847847 I would like to know how to compute the length of each SEQUENCE. Maybe the following code? data var1 var2 1 This is a string 12 2 This is another string 34 nchar(data[,1]) [1] 16 22 Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] length of a string
Hi, sapply(levels(df[,SEQUENCE]), nchar) Where 'df' is your data.frame -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 05/09/07, João Fadista [EMAIL PROTECTED] wrote: Dear all, I would like to know how can I compute the length of a string in a dataframe. Example: SEQUENCE ID TGCTCCCATCTCCACGGHR04FS00645 ACTGAACTCCCATCTCCAAT HR0595847847 I would like to know how to compute the length of each SEQUENCE. Best regards, João Fadista [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] length of a string
On 05-Sep-07 13:50:57, João Fadista wrote: Dear all, I would like to know how can I compute the length of a string in a dataframe. Example: SEQUENCE ID TGCTCCCATCTCCACGGHR04FS00645 ACTGAACTCCCATCTCCAAT HR0595847847 I would like to know how to compute the length of each SEQUENCE. Best regards, João Fadista nchar(ACTGAACTCCCATCTCCAAT) [1] 20 seems to work. Find it, and related functions, with help.search(character) As it happens, help.search(string) will not help! Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 05-Sep-07 Time: 15:05:22 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] geotiff or tiff files with world files
Hi, I have a matrix of data which i can vizualize as an image - for example. I would like to save this image as a geotiff file or at a tiff file with a world file which holds the projection of my data (ultimately the data represent a map of some sort). I know i can save the data as an ESRI grid, but i am not interested in that. I wonder if anybody knows about any code which will help me do that. Thanks in advance, Monica _ s. It's easy! aspxmkt=en-us [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] length of a string
As long as you keep in mind Prof. Ripley's comment, you're going to be fine with nchar(). http://tolstoy.newcastle.edu.au/R/e2/devel/07/05/3450.html Remember that what you want exactly is given by nchar(obj, type=chars), which is **NOT** the default on R 2.5.1 (only on R-2.6.0). In your particular situation, assuming R-2.5.1, nchar(obj) works, but i'm afraid it's only a coincidence. b On Sep 5, 2007, at 11:05 AM, (Ted Harding) wrote: On 05-Sep-07 13:50:57, João Fadista wrote: Dear all, I would like to know how can I compute the length of a string in a dataframe. Example: SEQUENCE ID TGCTCCCATCTCCACGGHR04FS00645 ACTGAACTCCCATCTCCAAT HR0595847847 I would like to know how to compute the length of each SEQUENCE. Best regards, João Fadista nchar(ACTGAACTCCCATCTCCAAT) [1] 20 seems to work. Find it, and related functions, with help.search(character) As it happens, help.search(string) will not help! Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 05-Sep-07 Time: 15:05:22 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] multivariate ARMA simulation
Dear Sirs: I am trying to simulate a ARMA multivariate model using AR-array(c(1,0.5,0,0.1,0,0.4,1,0.8),c(2,2,2)) mod- ARMA(A=AR, B=diag(1,2),C=NULL) y-simulate(mod,sampleT=100) in the package dse1, but how can I specify the covariance matrix for the noise of the model? and does the above procedure use Gaussian white noise?? Thank you Hanwen Zhang - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Multiple xyplots
Hi everyone, I'm hoping you can give me some pointers. I have a requirement to draw multiple (103) xy line plots onto one output device. Ideally the plots should be displayed in a hexagonal grid (example at www.maladmin.com/example.jpg). I can calculate the locations for each waveform but am wondering how to create multiple plotting areas. I have come accross references to a package grid (which doesn't seem to be in my CRAN mirror probability.ca) and lattice but I'm not sure if I'm on the correct lines. Any advice gratefully received. Thanks Tom -- --Tom Wright | Contact me: | | Skype: 0121 288 0756tomwright01) | | MSN: [EMAIL PROTECTED]| | Jabber: [EMAIL PROTECTED] | | ICQ: 423913453| |___| Ever since prehistoric times, wise men have tried to understand what, exactly, make people laugh. That's why they were called wise men. All the other prehistoric people were out puncturing each other with spears, and the wise men were back in the cave saying: How about: Would you please take my wife? No. How about: Here is my wife, please take her right now. No How about: Would you like to take something? My wife is available. No. How about ... -- Dave Barry, Why Humor is Funny __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Q: selecting a name when it is known as a string
D. R. Evans said the following at 09/04/2007 04:14 PM : I am 100% certain that there is an easy way to do this, but after I have reconsidered this and now believe it to be essentially impossible (or at the very least remarkably difficult) although I don't understand why it is so :-( At least, I spent another two hours trying variations on the suggestions I received, but still nothing worked properly. It sure seems like it _ought_ to be easy, because of the following argument: If I type an expression such as A - something then R is perfectly capable of parsing the something and executing it and assigning the result to A. So it seems to follow that it ought to be able to parse a string that contains exactly the same sequence of characters (after all, why should the R parsing engine care whether the input string comes from the terminal or from a variable?) and therefore it should be possible to assign something to a variable and then have R parse that variable precisely as if it had been typed. That was my logic as to why this ought to be easy, anyway. (And there was the subsidiary argument that this is easy in the other languages I use, but R is sufficiently different that I'm not certain that that argument carries much force.) It does seem that there are several ways to make the lo - loess(percent ~ ncms * ds, d, control=loess.control(trace.hat = 'approximate')) command work OK if the right hand side is in a character variable, but I haven't been able to find a way to make grid - data.frame(expand.grid(ds=MINVAL:MAXVAL, ncms=MINCMS:MAXCMS)) work. I always end up with a parse error or a complaint that 'newdata' does not contain the variables needed when I perform the next task: plo - predict(lo, grid). So I guess I have to stick with half a dozen compound if statements, all of which do essentially the same thing :-( __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bootstrap confidence intervals with previously existing bootstrap sample
On Tue, 4 Sep 2007, [EMAIL PROTECTED] wrote: I am new to R. I would like to calculate bootstrap confidence intervals using the BCa method for a parameter of interest. My situation is this: I already have a set of 1000 bootstrap replicates created from my original data set. I have already calculated the statistic of interest for each bootstrap replicate, and have also calculated the mean for this statistic across all the replicates. Now I would like to calculate Bca confidence intervals for this statistic. Is there a way to import my previously-calculated set of 1000 statistics into R, and then calculate bootstrap confidence intervals around the mean from this imported data? I have found the code for boot.ci in the manual for the boot package, but it looks like it requires that I first use the boot function, and then apply the output to boot.ci. Because my bootstrap samples already exist, I don't want to use boot, but just want to import the 1000 values I have already calculated, and then get R to calculate the mean and Bca confidence intervals based on these values. Is this possible? Brian Ripley wrote: Yes, it is possible but you will have to study the internal structure of an object of class boot (which is documented on the help page) and mimic it. You haven't told us which type of bootstrap you used, which is one of the details you need to supply. It might be slightly easier to work with function bcanon in package bootstrap, which you would need to edit to suit your purposes. I don't know why you have picked on the BCa method: my experience is that if you need to correct the basic method you often need far more than 1000 samples to get reliable results. You can do the BCa, but you need to supply parameters: z0: typically calculated from the fraction of bootstrap statistics that are = the original statistic acceleration: based on the skewness of the empirical influence function, typically calculated using the jackknife I agree that you should do far more than 1000 samples. The BCa uses bootstrap quantiles that are adjusted based on the z0 and acceleration parameters, and estimating z0 from the bootstrap samples magnifies the Monte Carlo error. You need roughly double as many bootstrap samples as for the bootstrap percentile interval; e.g. 10^4 instead of 5000. If computational expense is an issue, you might prefer bootstrap tilting intervals, which require about 1/37 as many bootstrap samples as the BCa for comparable Monte Carlo variability. Quick overview of confidence intervals: accuracycomments t intervals 1/sqrt(n) Using either formula or bootstrap standard error; poor in the presence of skewness. bootstrap percentile1/sqrt(n) Good quick-and-dirty procedure. Partial skewness correction. Poor if the statistic is biased. bootstrap t 1/n Good coverage, but interval width can vary wildly when n is small. BCa 1/n Current best overall, but you need a lot of bootstrap samples, e.g. 10^4. tilting 1/n Low Monte Carlo variability, so can use fewer bootstrap samples. Difficult to implement, and requires that statistic can be calculated with weights. Advertisement 1: tilting is available in S+Resample, available free from www.insightful.com/downloads/libraries Advertisement 2: I talk about these more in my short course, Bootstrap Methods and Permutation Tests Oct 10-11 San Francisco, 3-4 Oct UK. http://www.insightful.com/services/training.asp | Tim Hesterberg Senior Research Scientist | | [EMAIL PROTECTED] Insightful Corp.| | (206)802-23191700 Westlake Ave. N, Suite 500 | | (206)283-8691 (fax) Seattle, WA 98109-3044, U.S.A. | | www.insightful.com/Hesterberg | __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to obtain intercept statistics in anova withwithin-subject factors?
Sorry, I did not think about the nested design (did not read carefully enough).
Another thing to try (untested) is to create a column of 1's and include that
specifically and exclude the default intercept, then your column of 1's acts as
the intercept, but a p-value is returned from it.
Note that sometimes the other terms may change their definition without an
intercept. I had success at one time (quite a while ago, so things could have
changed) with specifying the formula as:O
Y ~ 1 + x1 + x2 + x3 -1
Which caused it to include the intercept, use that info when setting the dummy
vars for x1-x3, then removed the intercept.
Again this is all untested, just an idea to try.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Sid Kouider
Sent: Friday, August 31, 2007 3:19 PM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] How to obtain intercept statistics in anova
withwithin-subject factors?
Dear Greg,
Thanks very much for you advice.
Unfortunately, although summary.lm applied on aov objects
indeed shows the intercept's statistics, this function does
not (seem to) work with within-participant designs. As soon
as I enter the info on the error term (see the example in my
first message), then summary.lm(aov_object) crashes (Error
in if (p == 0) { : argument is of length zero).
-Sid
Subject: RE: [R] (no subject)
Date: Thu, 30 Aug 2007 13:43:19 -0600
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; r-help@stat.math.ethz.ch
Try calling summary.lm on your object (if it is an aov object then
summary
calls summary.aov which does not show the intercept, but
calling summary.lm directly does give info on the intercept).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Message d'origine-
De : Sid Kouider [mailto:[EMAIL PROTECTED] Envoyé :
jeudi 30 août
2007 16:54 À : '[EMAIL PROTECTED]'
Objet : How to obtain intercept statistics in anova with
within-subject factors?
Dear R users,
I am looking for an easy (i.e., direct) way of obtaining
the F and p
values from the intercept in anovas with within-subject designs.
My data are from a psychophysics experiment where I am using d' (d-
prime) values obtained from 3 modalities of presentation in each
subject. I would like to know not only whether there is an
effect of
modality, but also whether the main effect is significant (meaning
that d' 0).
I know that a t.test again the null mean would provide me
with similar
stats on the main effect, but I would like to get those
stats in an F
form, for consistency with the stats on the other factors
of interest.
As far as I understand how R works, the function anova
provides you
with such information but it is restricted to between-group
analyses.
For within-subject designs, one has to use summary(aov)
but stats on
the intercept are not included in the result of this
function. I have
pasted an example below. As one can see, only the Sum of Sq
and Mean
Sq are given for the main effect.
Thank you for any advice you can provide, -Sid
summary(aov(x~mod+Error(suj/(mod)), data=dp))
Error: suj Df Sum Sq Mean Sq F value Pr(F)
Residuals 10 19.5977 1.9598
Error: suj:mod
Df Sum Sq Mean Sq F value Pr(F) mod
2 8.2475 4.1237 4.2955 0.02806 *
Residuals 20 19.2005 0.9600
---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1
' ' 1 Error:
Within
Df Sum Sq Mean Sq F value Pr(F)Residuals 33 55.812 1.691
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Re: [R] Multiple xyplots
On 9/5/07, Tom Wright [EMAIL PROTECTED] wrote: Hi everyone, I'm hoping you can give me some pointers. I have a requirement to draw multiple (103) xy line plots onto one output device. Ideally the plots should be displayed in a hexagonal grid (example at www.maladmin.com/example.jpg). I can calculate the locations for each waveform but am wondering how to create multiple plotting areas. I have come accross references to a package grid (which doesn't seem to be in my CRAN mirror probability.ca) and lattice but I'm not sure if I'm on the correct lines. Any advice gratefully received. grid seems like the right choice to me (it comes bundled with R, so it's not available as a separate package). Here's an example that may give you a few hints: library(grid) grid.newpage() for (i in 1:30) grid.lines(x = 0:20/21, y = sin(70 * runif(1) * 0:20/21), vp = viewport(x = runif(1), y = runif(1), height = 0.05, width = 0.05)) See the package documentation, or Paul Murrell's book R Graphics for more. -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Monotone splines
Hello, i have a little problem with R and i hope you can help me. I want to use splines to estimate a function but i want to force the interpolation to be monotone. Is this possible with R ? Thank you, Rémi. - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] question about non-linear least squares in R
Here is one way of getting a reasonable fit: 1. Scale your y's by dividing all values by 1e6. 2. Plot x vs. y. The plot looks like a quadratic function. 3. Fit a quadratic const. + B*x^2 - this a linear regression problem so use lm. 4. Plot the predictions. 5. Eyball the necessary shift - MA is around 0.01. Refit const. + B*(x-.01)^2. Should get const.=1.147 and B=139.144 6. Use start=list(const.= 1.147, A=0, B=1.147, MA=.01). nls should converge in 4 iterations. In general, good starting points may be crucial to nls convergence. Scaling the y's to reasonable values also helps. Hope this helps, Andy __ Andy Jaworski 518-1-01 Process Laboratory 3M Corporate Research Laboratory - E-mail: [EMAIL PROTECTED] Tel: (651) 733-6092 Fax: (651) 736-3122 Yu (Warren) Wang [EMAIL PROTECTED] To Sent by: r-help@stat.math.ethz.ch [EMAIL PROTECTED] r-help@stat.math.ethz.ch at.math.ethz.chcc Subject 09/05/2007 02:51 [R] question about non-linear least AMsquares in R Hi, everyone, My question is: It's not every time that you can get a converged result from the nls function. Is there any solution for me to get a reasonable result? For example: x - c(-0.06,-0.04,-0.025,-0.015,-0.005,0.005,0.015,0.025,0.04,0.06) y - c(1866760,1457870,1314960,1250560,1184850,1144920,1158850,1199910,1263850,1452520) fitOup- nls(y ~ constant + A*(x-MA)^4 + B*(x-MA)^2, start=list(constant=1000, A=1, B=-100, MA=0), control=nls.control(maxiter=100, minFactor=1/4096), trace=TRUE) For this one, I cannot get the converged result, how can I reach it? To use another funtion or to modify some settings for nls? Thank you very much! Yours, Warren __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ecological meaning of randomForest vegetation classification?
Hi, everyone, I haven't found anything similar in the forum, so here's my problem (I'm no expert in R nor statistics): I have a data set of 59.000 cases with 9 variables each (fractional coverage of 9 different plant types, such as deciduous broad-leaved temperate trees or evergreen tropical trees etc.), which was generated by a vegetation model. In order to evaluate the quality of the vegetation model's output, I want to compare it to a land-cover data set which has 23 different land-cover types (such as needle leaved evergreen forest, dense broad-leaved forest, barren, etc.). A statistician advised me to use the randomForest package in R and using a sub-set to generate the random Forest, I get a very good prediction for the rest. However, I need to evaluate how meaningful this classification is in an ecological sense (boreal trees should not play a role in the definition of tropical land-cover types, for example), otherwise I cannot judge the quality of the vegetation model's output. Unfortunately, randomForest gives me about 15.000 splits of which about 5000 are end branches (rough guess), so it's very hard and time-consuming to check each single branch of one of the final trees for its ecological meaning. Is there any utility to summarize the characteristics of each of the 23 prediction classes? Such as land-cover class 1 has less than 5% of plant types 1-5, 20-50% of plant type 7 and at least 30% of plant type 8. Or is there a more suitable method to classify my data? Thanks a lot in advance! Christoph Click on the following link for the Netherlands Environmental Assessment Agency(MNP)mission and contact information: http://www.mnp.nl/signature.html Klik op de volgende link voor missie en contactinformatie van het Milieu- en Natuurplanbureau (MNP): http://www.mnp.nl/signature.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] kendalls tau c
Barbara Diane-Spillmann wrote: dear all, does anybody know if cor.test with method=kendall calculates kendalls tau a b or c? I need to get p values for kendalls tau c... May I ask what the difference between Kendall's tau a, b and c is? Any references? Uwe Ligges thank you very much for any kind of hint. Barbara __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Single plot multiple levels in x?
Greg,
This is great, exactly what I was looking for.
Thanks.
Rich
-Original Message-
From: Greg Snow [EMAIL PROTECTED]
Sent: Aug 31, 2007 2:55 PM
To: Richard Yanicky [EMAIL PROTECTED], Uwe Ligges [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Subject: RE: [R] Single plot multiple levels in x?
Try this code (with the mydf that you generate below):
library(TeachingDemos)
plot( c(0,5), c(0,1), xlab='State', ylab='ylab', axes=FALSE, type='n' )
axis(1, at= (1:5) - 0.5, labels=paste('state',1:5))
box()
for(i in 1:5){
with( subset(mydf, State==i),
subplot( plot(Position, `Pct Recurr`, col=i,
yaxt=ifelse( i==1, 's', 'n' ), ylab='',xlab='',
cex.axis=0.5),
x=c(i-1,i), y=c(0,1) )
)
}
Is that close to what you want?
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Richard Yanicky
Sent: Thursday, August 30, 2007 11:37 AM
To: Uwe Ligges
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Single plot multiple levels in x?
Uwe,
Here is some code to create some data then a plot (The plot
was done in another package). The plot is included only to
reference the structure of the x-axis. I can't get R to do
something similar.
State - seq (1:5);
posi - seq (0.5,62525,199.233)
mydf-NULL;
for ( i in 1:5) {
df1-data.frame(i,posi);
mydf - rbind(mydf,df1); }
myy-rep(-100.01:100.01,length=nrow(mydf));
mydf-cbind(mydf,myy);
names(mydf) - c(State,Position,Pct Recurr);
I would like to somehow:
plot(c(mydf[,1],mydf[,2]),mydf[,3]) and end up with the
nested structure on the x-axis.
Thanks,
Richard
-Original Message-
From: Uwe Ligges [EMAIL PROTECTED]
Sent: Aug 30, 2007 10:50 AM
To: Richard Yanicky [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Single plot multiple levels in x?
Richard Yanicky wrote:
Uwe,
I have looked into lattice and can't seem to make this
work. I can easily make multiple panels but this isn't what I
am looking to do. Any suggestions on which functions to use?
the axis function seems a natural place to start but I still
can't seem to make it happen.
If lattice is not what you want, I do not understand what
you mean. Can
you give a more elaborated example, please?
Uwe
HELP!
Richard
-Original Message-
From: Uwe Ligges [EMAIL PROTECTED]
Sent: Aug 30, 2007 5:59 AM
To: Richard Yanicky [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Single plot multiple levels in x?
Richard Yanicky wrote:
Plotting with 2 x axis?
One axis inside another, for example salary within state,
1-50 | 50 - 100 | 100+ | 1- 50 | 50 -100 |
100+ | ... repeated bins for salary
AL !
AR .. more states
Sounds like the lattice package does exactly what you want, but
without any reproducible example.
Uwe Ligges
The values are all stored with a single data frame. I have tried
different things with the axis function and done many
searches for
plotting. Can't find a direct reference
Thanks.
Richard
__
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and provide commented, minimal, self-contained,
reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] level names with groupdedData object
How do I get a grouped data object to use the level names from the input data set? first I gave the levels some names like this: male -factor(male) levels(male) - c(“Girls”,”Boys”) Then I created a groupdedData object but the male variable in not part of the grouping formula. Then I fit an lme and then use this code to create a plot: plot(myfit.lme, resid(. , type=”p”) ~ fitted(.) | male , id=0.05, adj=-0.3 ) But the plot just used “Male” in both panels. How do I get the plot to used Girls Boys? Thanks, Dean __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] element wise opertation between a vector and a list
?mapply
mapply('+', a, b, SIMPLIFY=FALSE)
colSums(mapply('+', a, b))
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Yongwan Chun
Sent: Monday, September 03, 2007 12:36 AM
To: r-help@stat.math.ethz.ch
Subject: [R] element wise opertation between a vector and a list
I want to try to get a result of element wise addition
between a vector and a list. It can be done with for
statement. In order to reducing computing time, I have tried
to avoid for state. If anybody give me an idea, I would
apprecite it much.
for example, with a b as below lines,
a- list(c(1,3),c(1,2),c(2,3))
b-c(10,20,30)
I would like to have a list (like d) or a vector (like e)
as below.
d-list(c((1+10),(3+10)),c((1+20),(2+20)),c((2+30),(3+30)))
e- c((1+10)+(3+10),(1+20)+(2+20),(2+30)+(3+30))
Thanks,
Yongwan Chun
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[R] list element to matrix
I have created a list of matrices using sapply or lapply and wish to extract each of the matrices as a matrix. Some of them are 2x2, 3x3, etc. I can do this one at a time as: M1-as.matrix(D[[1]]) How can repeat this process for an unknown number of entries in the list? In other words, how shall I index M1? Diana __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ecological meaning of randomForest vegetation classification? [Broadcast]
Hi Christoph,
I'm not exactly sure what you're looking for, but I'll take a stab
anyway.
The trees in a random forest is not designed to be interpreted as one
would
with an ordinary tree. There are several things you may try to see if
they help you any. One is the distribution of votes. It looks like you
are
classifying each data point into one of many possible classes. RF with
give
you the fraction of trees in the forest that classified the observation
as
a particular class (and the class with the highest fraction of votes is
the
predicted class). Another is the partial dependence plot: You can
use
plot(importance(rf.object)) to see which variables are the most
important,
and then use partialPlot() to examine their marginal effects. These
offer
some clue of what the RF black box is doing, and hopefully will make
some
sense to you.
Best,
Andy
From: Christoph Muller
Hi, everyone,
I haven't found anything similar in the forum, so here's my
problem (I'm no
expert in R nor statistics):
I have a data set of 59.000 cases with 9 variables each (fractional
coverage of 9 different plant types, such as deciduous broad-leaved
temperate trees or evergreen tropical trees etc.), which was
generated by a
vegetation model.
In order to evaluate the quality of the vegetation model's
output, I want
to compare it to a land-cover data set which has 23 different
land-cover
types (such as needle leaved evergreen forest, dense
broad-leaved forest,
barren, etc.).
A statistician advised me to use the randomForest package in
R and using a
sub-set to generate the random Forest, I get a very good
prediction for the
rest.
However, I need to evaluate how meaningful this
classification is in an
ecological sense (boreal trees should not play a role in the
definition of
tropical land-cover types, for example), otherwise I cannot judge the
quality of the vegetation model's output.
Unfortunately, randomForest gives me about 15.000 splits of
which about
5000 are end branches (rough guess), so it's very hard and
time-consuming
to check each single branch of one of the final trees for its
ecological
meaning.
Is there any utility to summarize the characteristics of each
of the 23
prediction classes? Such as land-cover class 1 has less than
5% of plant
types 1-5, 20-50% of plant type 7 and at least 30% of plant type 8.
Or is there a more suitable method to classify my data?
Thanks a lot in advance!
Christoph
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Assessment
Agency(MNP)mission and contact information:
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[R] Lattice: key with expression function
HI all, I'm trying (unsuccessfully) to add the degree symbol to each line of text in my legend (within xyplot). Here is the line of code, which fails to interpret the expression function: auto.key =list(points = FALSE,text=paste(levels(as.factor(divertSST2$temp)),expression(degree)). ..), I just get: 7 degree 8 degree 9 degree If I place 'expression' outside or just after the paste function it also doesn't work. Any suggestions are well received! Thanks Michael Folkes [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rug() colors
You could make multiple calls to the rug function using a different color for each call. Using tapply or by may automate this for you. Another option would be to create your own rug using the segments function (which will plot multiple colors). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Tuesday, September 04, 2007 12:00 PM To: r-help@stat.math.ethz.ch Subject: [R] rug() colors Hello, I have a simple question on rug(). Currently there is only one color possible for the rug. Is it possible to plot a the rug with different colors, for each rug item ? Thx. Bjoern __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice: key with expression function
On 9/5/07, Folkes, Michael [EMAIL PROTECTED] wrote: HI all, I'm trying (unsuccessfully) to add the degree symbol to each line of text in my legend (within xyplot). Here is the line of code, which fails to interpret the expression function: auto.key =list(points = FALSE,text=paste(levels(as.factor(divertSST2$temp)),expression(degree)). ..), I just get: 7 degree 8 degree 9 degree That's because paste(foo, expression(degree)) [1] foo degree If I place 'expression' outside or just after the paste function it also doesn't work. auto.key = list(text = expression(paste(foo, degree))) should work. I think the problem is that you want a vector of expressions, and that's a bit harder to get. I'm not sure what the best solution is, but if everything else fails, you could try using parse(text=) -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Q: selecting a name when it is known as a string
If your main goal is to do a loess fit, then make predictions from that,
then using the 'get' function may do what you want:
tmp.var - get(ORDINATE)
lo - loess(percent ~ ncms * tmp.var, d, ...
grid - expand.grid(tmp.var=MINVAL:MAXVAL, ncms=MINCMS:MAXCMS)
predict(lo, grid)
Here you stick with the name tmp.var so it matches between the formula
and the data frame, the predictions will be what you want. If you also
want to print out some of the model summary then realize that it will
have tmp.var in place of ds or whatever (It may not be too much trouble
to change that name after the fact if that is what is important).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of D. R. Evans
Sent: Tuesday, September 04, 2007 4:15 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Q: selecting a name when it is known as a string
I am 100% certain that there is an easy way to do this, but
after experimenting off and on for a couple of days, and
searching everywhere I could think of, I haven't been able to
find the trick.
I have this piece of code:
...
attach(d)
if (ORDINATE == 'ds')
{ lo - loess(percent ~ ncms * ds, d,
control=loess.control(trace.hat =
'approximate'))
grid - data.frame(expand.grid(ds=MINVAL:MAXVAL,
ncms=MINCMS:MAXCMS)) ...
then there several almost-identical if statements for
different values of ORDINATE. For example, the next if
statement starts with:
...
if (ORDINATE == 'dsl')
{ lo - loess(percent ~ ncms * dsl, d,
control=loess.control(trace.hat =
'approximate'))
grid - data.frame(expand.grid(dsl=MINVAL:MAXVAL,
ncms=MINCMS:MAXCMS)) ...
This is obviously pretty silly code (although of course it does work).
I imagine that my question is obvious: given that I have a
variable, ORDINATE, whose value is a string, how do I
re-write statements such as the lo - and grid -
statements above so that they use ORDINATE instead of the
hard-coded names ds and dsl.
I am almost sure (almost) that it has something to do with
deparse(), but I couldn't find the right incantation, and
the ?deparse() help left my head swimming.
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Re: [R] Multiple xyplots
Take a look at the my.symbols function in the TeachingDemos package. The last example shows how to create a hexagonal grid and the 2nd to last example shows how to plot several small line plots onto a larger plot. Combining these 2 examples should give you what you want. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Tom Wright Sent: Wednesday, September 05, 2007 8:54 AM To: r-help@stat.math.ethz.ch Subject: [R] Multiple xyplots Hi everyone, I'm hoping you can give me some pointers. I have a requirement to draw multiple (103) xy line plots onto one output device. Ideally the plots should be displayed in a hexagonal grid (example at www.maladmin.com/example.jpg). I can calculate the locations for each waveform but am wondering how to create multiple plotting areas. I have come accross references to a package grid (which doesn't seem to be in my CRAN mirror probability.ca) and lattice but I'm not sure if I'm on the correct lines. Any advice gratefully received. Thanks Tom -- --Tom Wright | Contact me: | | Skype: 0121 288 0756tomwright01) | | MSN: [EMAIL PROTECTED]| | Jabber: [EMAIL PROTECTED] | | ICQ: 423913453| |___| Ever since prehistoric times, wise men have tried to understand what, exactly, make people laugh. That's why they were called wise men. All the other prehistoric people were out puncturing each other with spears, and the wise men were back in the cave saying: How about: Would you please take my wife? No. How about: Here is my wife, please take her right now. No How about: Would you like to take something? My wife is available. No. How about ... -- Dave Barry, Why Humor is Funny __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] list element to matrix
You get the number of list elements with length(D), the dimensions of M1 with dim(M1) see help with: ?dim ?length Hope this helps... I have created a list of matrices using sapply or lapply and wish to extract each of the matrices as a matrix. Some of them are 2x2, 3x3, etc. I can do this one at a time as: M1-as.matrix(D[[1]]) How can repeat this process for an unknown number of entries in the list? In other words, how shall I index M1? Diana -- Friedrich Schuster [EMAIL PROTECTED] Tel.: +49 6221 737474 Tel.: +49 163 7374744 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Q: selecting a name when it is known as a string
For the column names of the result of expand.grid(), I would just assign them the values I wanted, like this: x - expand.grid(tmp=1:3,y=1:2) x tmp y 1 1 1 2 2 1 3 3 1 4 1 2 5 2 2 6 3 2 colnames(x)[1] - whatever x whatever y 11 1 22 1 33 1 41 2 52 2 63 2 -- Tony Plate D. R. Evans wrote: D. R. Evans said the following at 09/04/2007 04:14 PM : I am 100% certain that there is an easy way to do this, but after I have reconsidered this and now believe it to be essentially impossible (or at the very least remarkably difficult) although I don't understand why it is so :-( At least, I spent another two hours trying variations on the suggestions I received, but still nothing worked properly. It sure seems like it _ought_ to be easy, because of the following argument: If I type an expression such as A - something then R is perfectly capable of parsing the something and executing it and assigning the result to A. So it seems to follow that it ought to be able to parse a string that contains exactly the same sequence of characters (after all, why should the R parsing engine care whether the input string comes from the terminal or from a variable?) and therefore it should be possible to assign something to a variable and then have R parse that variable precisely as if it had been typed. That was my logic as to why this ought to be easy, anyway. (And there was the subsidiary argument that this is easy in the other languages I use, but R is sufficiently different that I'm not certain that that argument carries much force.) It does seem that there are several ways to make the lo - loess(percent ~ ncms * ds, d, control=loess.control(trace.hat = 'approximate')) command work OK if the right hand side is in a character variable, but I haven't been able to find a way to make grid - data.frame(expand.grid(ds=MINVAL:MAXVAL, ncms=MINCMS:MAXCMS)) work. I always end up with a parse error or a complaint that 'newdata' does not contain the variables needed when I perform the next task: plo - predict(lo, grid). So I guess I have to stick with half a dozen compound if statements, all of which do essentially the same thing :-( __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Round Off Error Variance
I am trying to specify a round off error distribution, but I am not sure what the variance of round off error would be for R. Does anyone have a suggestion? Thanks Todd Remund (435) 863-8172 Science Engineering [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question about making array dataset inside a package
Hello I have a question about how I can include an array dataset inside a package. In our package, one R function is designed to take an array of 2 rows * 2 columns * k levels as input data enter by the user, where k is positive integers. I am trying to include a 3-D array of 2-by-2-by-8 as dataset in the package, so users can load the data using data(dataname). We prefer to load the data as dataset, rather than include the long syntax in the help file for users to copy and paste. I can generate array in R console using long chain of syntax like:arraydat=array(.omitted...) However, I cannot figure a way to save the data in 3D array data frame format using write(), write.table(), or use data.frame() etc. If I directly copy-paste the screen output to a text file. I cannot read it into R using like: arraydat=read.table(array.txt, header=TRUE) This will give me a 2 rows by (2*k) columns two-dimension data set, and lost all level (or depth) structure for the purpose of our R function. The header is messed too. I appreciate if any expert can tell me how to include/save array stucture in dataset in R, and also how to read it into R to preserve the header and array structure. Thank you for your time. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about making array dataset inside a package
Please read
http://cran.r-project.org/doc/manuals/R-exts.html
especially Section 1.1.3. Use save to create an Rdata file.
Max
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Wallace Hui
Sent: Wednesday, September 05, 2007 4:18 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Question about making array dataset inside a package
Hello
I have a question about how I can include an array dataset inside a
package. In our package, one R function is designed to take an array
of 2 rows * 2 columns * k levels as input data enter by the user,
where k is positive integers. I am trying to include a 3-D array of
2-by-2-by-8 as dataset in the package, so users can load the data
using data(dataname). We prefer to load the data as dataset, rather
than include the long syntax in the help file for users to copy and
paste. I can generate array in R console using long chain of syntax
like:arraydat=array(.omitted...)
However, I cannot figure a way to save the data in 3D array data frame
format
using write(), write.table(), or use data.frame() etc. If I directly
copy-paste the screen output to a text file. I cannot read it into R
using like:
arraydat=read.table(array.txt, header=TRUE)
This will give me a 2 rows by (2*k) columns two-dimension data set,
and lost all level (or depth) structure for the purpose of our R
function. The header is messed too.
I appreciate if any expert can tell me how to include/save array
stucture in dataset in R, and also how to read it into R to preserve
the header and array structure.
Thank you for your time.
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Re: [R] Lattice: key with expression function
Thanks to Deepayan...again, for suggesting parse. Here is how I added the degree symbol to a vector of text for my xyplot legend: auto.key =list(points = FALSE,text=parse(text = paste(levels(as.factor(divertSST2$temp)), *degree, sep = ))), For me the tricky part was learning about adding the '*'. I found that in this suggestion: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/78961.html Michael Folkes -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: September 5, 2007 11:27 AM To: Folkes, Michael Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Lattice: key with expression function On 9/5/07, Folkes, Michael [EMAIL PROTECTED] wrote: HI all, I'm trying (unsuccessfully) to add the degree symbol to each line of text in my legend (within xyplot). Here is the line of code, which fails to interpret the expression function: auto.key =list(points = FALSE,text=paste(levels(as.factor(divertSST2$temp)),expression(degree) ). ..), I just get: 7 degree 8 degree 9 degree That's because paste(foo, expression(degree)) [1] foo degree If I place 'expression' outside or just after the paste function it also doesn't work. auto.key = list(text = expression(paste(foo, degree))) should work. I think the problem is that you want a vector of expressions, and that's a bit harder to get. I'm not sure what the best solution is, but if everything else fails, you could try using parse(text=) -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] list element to matrix
If they are already a matrix in the list, then you don't have to use 'as.matrix'; you can just say: M1 - D[[1]] Now the question is, what do you mean by how do you index M1? Do you want to go through the list applying a function to each matrix? If so, then just 'lapply'. For example, to get the column means, you would do: mean.list - lapply(D, colMeans) Can you explain in a little more detail the problem you are trying to solve. On 9/5/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: I have created a list of matrices using sapply or lapply and wish to extract each of the matrices as a matrix. Some of them are 2x2, 3x3, etc. I can do this one at a time as: M1-as.matrix(D[[1]]) How can repeat this process for an unknown number of entries in the list? In other words, how shall I index M1? Diana __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] writing elements in list as a data frame
Dear R-helpers, Lists in R are stumbling block for me. I kindly ask you to help me able to write a data-frame. I have a list of lists. sls[1:2] $Andromeda_maya1 x y [1,] 369 103 [2,] 382 265 [3,] 317 471 [4,] 169 465 [5,] 577 333 $Andromeda_maya2 x y [1,] 173 507 [2,] 540 395 [3,] 268 143 [4,] 346 175 [5,] 489 91 I want to be able to write a data.frame like the following: X Y Name 369 103 Andromeda_maya1 382 265 Andromeda_maya1 317 471 Andromeda_maya1 169 465 Andromeda_maya1 577 333 Andromeda_maya1 173 507 Andromeda_maya2 540 395 Andromeda_maya2 268 143 Andromeda_maya2 346 175 Andromeda_maya2 489 91 Andromeda_maya2 Is there a way to convert this list-of-list into a data.frame. Thanks srini __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] writing elements in list as a data frame
Try this:
sls - list(a=matrix(sample(10), ncol=2, dimnames=list(NULL, c('x', 'y'))),
+ b=matrix(sample(16), ncol=2, dimnames=list(NULL, c('x', 'y'
sls
$a
x y
[1,] 8 2
[2,] 9 10
[3,] 4 1
[4,] 5 7
[5,] 3 6
$b
x y
[1,] 4 14
[2,] 3 15
[3,] 16 5
[4,] 1 9
[5,] 8 7
[6,] 10 2
[7,] 12 13
[8,] 11 6
# create output matrix
do.call('rbind', lapply(names(sls), function(.name){
+ data.frame(sls[[.name]], Name=.name)
+ }))
x y Name
1 8 2a
2 9 10a
3 4 1a
4 5 7a
5 3 6a
6 4 14b
7 3 15b
8 16 5b
9 1 9b
10 8 7b
11 10 2b
12 12 13b
13 11 6b
On 9/5/07, Srinivas Iyyer [EMAIL PROTECTED] wrote:
Dear R-helpers,
Lists in R are stumbling block for me.
I kindly ask you to help me able to write a
data-frame.
I have a list of lists.
sls[1:2]
$Andromeda_maya1
x y
[1,] 369 103
[2,] 382 265
[3,] 317 471
[4,] 169 465
[5,] 577 333
$Andromeda_maya2
x y
[1,] 173 507
[2,] 540 395
[3,] 268 143
[4,] 346 175
[5,] 489 91
I want to be able to write a data.frame like the
following:
X Y Name
369 103 Andromeda_maya1
382 265 Andromeda_maya1
317 471 Andromeda_maya1
169 465 Andromeda_maya1
577 333 Andromeda_maya1
173 507 Andromeda_maya2
540 395 Andromeda_maya2
268 143 Andromeda_maya2
346 175 Andromeda_maya2
489 91 Andromeda_maya2
Is there a way to convert this list-of-list into a
data.frame.
Thanks
srini
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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[R] confidence intervals of proportions from complex surveys
This is partly an R and partly a general statistics question. I'm trying to get confidence intervals of proportions (sometimes for subgroups) estimated from complex survey data. Because a function like prop.test() does not exist for the survey package I tried the following: 1) Define a survey object (PSU of clustered sample, population weights); 2) Use svyglm() of the package survey to estimate a binary logistic regression (family='binomial'): For the confidence interval of a single proportion regress the binary dependent variable on a constant (1), for confidence intervals of that variable for subgroups regress this variable on the groups (factor) variable; 3) Use predict() to obtain estimated logits and the respective standard errors (mod.dat specifiying either the constant or the subgroups): pred=predict(model,mod.dat,type='link',se.fit=T) and apply the following to obtain the proportion with its confidence intervals (for example, for conf.level=.95): lo.e = pred[1:length(pred)]-qnorm((1+conf.level)/2)*SE(pred) hi.e = pred[1:length(pred)]+qnorm((1+conf.level)/2)*SE(pred) prop = 1/(1+exp(-pred[1:length(pred)])) lo = 1/(1+exp(-lo.e)) hi = 1/(1+exp(-hi.e)) I think that in that way I get CI's based on asymptotic normality - either for a single proportion or split up into subgroups. Question: Is this a correct or a defensible procedure? Or should I use a different approach? Note that this approach should also allow to estimate CI's for proportions of subgroups taking into account the complex survey design. TIA, Dirk R version 2.5.1 Patched (2007-08-10 r42469) i386-pc-mingw32 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] question: randomization t-test function already defined in R?
Dear R Users, I am hoping you can help me. I have received code from a colleague who uses Matlab. I need to translate it into R. I am wondering if there is a randomization t-test (from non-parametric statistics) function already defined in R. (In Matlab the function is randtest.m.) ** QUESTION: Is anyone aware of a randomization t-test function in R? ** Thank you for your help, Karen --- Karen M. Green, Ph.D. Research Investigator Drug Design Group Sanofi Aventis Pharmaceuticals 1580 E. Hanley Blvd. Tucson, AZ 85737-9525 USA [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] question: randomization t-test function already defined in R?
On Wed, 5 Sep 2007, [EMAIL PROTECTED] wrote: Dear R Users, I am hoping you can help me. I have received code from a colleague who uses Matlab. I need to translate it into R. I am wondering if there is a randomization t-test (from non-parametric statistics) function already defined in R. (In Matlab the function is randtest.m.) I don't know what randtest in MATLAB does exactly, but I guess that you might want to look at the function independence_test() in package coin. The distribution argument should probably set to use either exact or approximate p values. hth, Z ** QUESTION: Is anyone aware of a randomization t-test function in R? ** Thank you for your help, Karen --- Karen M. Green, Ph.D. Research Investigator Drug Design Group Sanofi Aventis Pharmaceuticals 1580 E. Hanley Blvd. Tucson, AZ 85737-9525 USA [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] change all . to 0 in a data.frame
Hello, I read in a tab delimited text file via mydata = read.delim(myfile). The text file was originally an excel file where . was used in place of 0. Now all the columns which should be integers are factors. Any ideas how to change all the . to 0 and factors back to integer? Thanks a lot in advance for any suggestions, -- D - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] change all . to 0 in a data.frame
Here is one way. You might want to read in the data with 'as.is=TRUE' to prevent conversion to factors. x - data.frame(a=c(1,2,3,'.',5,'.')) str(x) 'data.frame': 6 obs. of 1 variable: $ a: Factor w/ 5 levels .,1,2,3,..: 2 3 4 1 5 1 # replace '.' with zero; either readin with 'as.is=TRUE' or convert to character x$a - as.character(x$a) x$a[x$a == '.'] - '0' x$a - as.numeric(x$a) str(x) 'data.frame': 6 obs. of 1 variable: $ a: num 1 2 3 0 5 0 On 9/5/07, Dieter Best [EMAIL PROTECTED] wrote: Hello, I read in a tab delimited text file via mydata = read.delim(myfile). The text file was originally an excel file where . was used in place of 0. Now all the columns which should be integers are factors. Any ideas how to change all the . to 0 and factors back to integer? Thanks a lot in advance for any suggestions, -- D - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] The variables combined in a table from other table and combination questions
Dear All: I need to have some data frame objects. First aa object: pH Formulation time Subject [1]1.2 F 0 1 [2]7.4 S 1 2 [3]MF 2 3 [4] 3 4 [5] ni Then, I need to produce 2*3(pH*formulation) different tables. This table includes column of (pH, Formulation, time S1 S2 S3 …..Si) and S1= subject 1, S2=subject 2 and so on. For example: bb1 table pH Formulation time S1 S2 S3….Si [1]1.2 F 0 [2] 1 [3] 2 [4] 3 [5] n For example: bb2 table pH Formulation time S1 S2 S3….Si [1]1.2 S 0 [2] 1 [3] 2 [4] 3 [5] n Moreover, the values of pH and Formulation column are the combination questions. The values of pH and Formulation column should be the combinations such as (1.2, F), (1.2, S), (1.2, MF), (7.4, F), (7.4, S), (7.4, MF) I am a beginner level in R and I have no idea how to do this. Could any one please help me. Thanks a lot!!! Best regrards Hsin Ya Lee __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Rggobi compilation error: display.c
On a ubuntu linux computer (Feisty, i386), I compile R and additional packages from source. The compiler is gcc 4.1.2. The problem is, I can run sudo R and successfully compile all packages (e.g., MASS, lattice) except rggobi. The error seems to be in display.c. My ggobi is in /usr/local/, which R can find. I don't think this is a dependence issue because install.packages(..., dependence=TRUE). The script complains about not finding /usr/local/lib/R/library/rggobi/libs/*, but that directory is there, with one file rggobi.so (possibly from an earlier successful compilation). Any suggestions on what I am doing wrong? Many thanks in advance. Yuelin. - R output --- install.packages(rggobi, repos = http://lib.stat.cmu.edu/R/CRAN;, dependencies = TRUE, clean = TRUE) trying URL 'http://lib.stat.cmu.edu/R/CRAN/src/contrib/rggobi_2.1.6.tar.gz' Content type 'application/x-gzip' length 424483 bytes opened URL == downloaded 414Kb * Installing *source* package 'rggobi' ... checking for pkg-config... /usr/bin/pkg-config checking pkg-config is at least version 0.9.0... yes checking for GGOBI... yes configure: creating ./config.status config.status: creating src/Makevars ** libs gcc -std=gnu99 -I/usr/local/lib/R/include -I/usr/local/lib/R/include -g -DUSE_EXT_PTR=1 -D_R_=1 -I/usr/local/include/ggobi -I/usr/include/gtk-2.0 -I/usr/include/libxml2 -I/usr/lib/gtk-2.0/include -I/usr/include/atk-1.0 -I/usr/include/cairo -I/usr/include/pango-1.0 -I/usr/include/glib-2.0 -I/usr/lib/glib-2.0/include -I/usr/include/freetype2 -I/usr/include/libpng12 -I/usr/local/include-fpic -g -O2 -c brush.c -o brush.o [... snipped ...] gcc -std=gnu99 -I/usr/local/lib/R/include -I/usr/local/lib/R/include -g -DUSE_EXT_PTR=1 -D_R_=1 -I/usr/local/include/ggobi -I/usr/include/gtk-2.0 -I/usr/include/libxml2 -I/usr/lib/gtk-2.0/include -I/usr/include/atk-1.0 -I/usr/include/cairo -I/usr/include/pango-1.0 -I/usr/include/glib-2.0 -I/usr/lib/glib-2.0/include -I/usr/include/freetype2 -I/usr/include/libpng12 -I/usr/local/include-fpic -g -O2 -c dataset.c -o dataset.o gcc -std=gnu99 -I/usr/local/lib/R/include -I/usr/local/lib/R/include -g -DUSE_EXT_PTR=1 -D_R_=1 -I/usr/local/include/ggobi -I/usr/include/gtk-2.0 -I/usr/include/libxml2 -I/usr/lib/gtk-2.0/include -I/usr/include/atk-1.0 -I/usr/include/cairo -I/usr/include/pango-1.0 -I/usr/include/glib-2.0 -I/usr/lib/glib-2.0/include -I/usr/include/freetype2 -I/usr/include/libpng12 -I/usr/local/include-fpic -g -O2 -c display.c -o display.o display.c: In function ‘RS_GGOBI_createDisplay’: display.c:37: warning: passing argument 3 of ‘klass-createWithVars’ makes pointer from integer without a cast display.c:37: warning: passing argument 4 of ‘klass-createWithVars’ from incompatible pointer type display.c:37: warning: passing argument 5 of ‘klass-createWithVars’ from incompatible pointer type display.c:37: error: too many arguments to function ‘klass-createWithVars’ display.c:39: warning: passing argument 4 of ‘klass-create’ from incompatible pointer type display.c:39: error: too many arguments to function ‘klass-create’ make: *** [display.o] Error 1 chmod: cannot access `/usr/local/lib/R/library/rggobi/libs/*': No such file or directory ERROR: compilation failed for package 'rggobi' ** Removing '/usr/local/lib/R/library/rggobi' ** Restoring previous '/usr/local/lib/R/library/rggobi' __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] element wise opertation between a vector and a list
On Wed, 5 Sep 2007, Greg Snow wrote:
?mapply
mapply('+', a, b, SIMPLIFY=FALSE)
colSums(mapply('+', a, b))
or
sapply( a, sum ) + b * sapply( a, length )
or even
sapply( a, sum ) + b * 2
if all list components in 'a' are of length 2.
Then there are the do.call( cbind , a ) incantations.
Chuck
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Yongwan Chun
Sent: Monday, September 03, 2007 12:36 AM
To: r-help@stat.math.ethz.ch
Subject: [R] element wise opertation between a vector and a list
I want to try to get a result of element wise addition
between a vector and a list. It can be done with for
statement. In order to reducing computing time, I have tried
to avoid for state. If anybody give me an idea, I would
apprecite it much.
for example, with a b as below lines,
a- list(c(1,3),c(1,2),c(2,3))
b-c(10,20,30)
I would like to have a list (like d) or a vector (like e)
as below.
d-list(c((1+10),(3+10)),c((1+20),(2+20)),c((2+30),(3+30)))
e- c((1+10)+(3+10),(1+20)+(2+20),(2+30)+(3+30))
Thanks,
Yongwan Chun
__
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Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED] UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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